# ComplexPlot3D and essential singularities [closed]

I am trying to understand the behavior of the functions

$$f(z) = e^\frac{1}{z} \qquad \text{and} \qquad \frac{1}{f(z)} = \frac{1}{e^\frac{1}{z}}$$

in the neighborhood of $$z=0$$.

From the power series representation of $$e^{1/z}$$, we know that $$f$$ has an essential singularity at $$z=0$$. Because $$1/f(z) = f(-z)$$ in this case, we know that $$1/f$$ is simply $$f$$ reflected about the origin, so $$1/f$$ should match the behavior of $$f$$ near $$0$$. In particular, $$1/f$$ also has an essential singularity at $$z=0$$.

In the newly released Mathematica 12, there is a function called ComplexPlot3D. Using the information on the documentation page, I was able to plot $$g(z) = z$$ and $$h(z)=1/z$$, as well as changing the plot range via the additional argument PlotRange → {0,100}.

This worked perfectly for $$z$$ and $$1/z$$ but only generated an empty plot with axes for $$e^\frac{1}{z}$$ or $$1/\left(e^\frac{1}{z}\right)$$.

I attempted to plot the same functions with different ranges using PlotRange but this did not change anything. This is the input I used:

 ComplexPlot3D[e^(1/z), {z, -1 - I, 1 + I}, PlotRange -> {0, 100}]


The problem shouldn't be the range, as $$e^\frac{1}{z}$$ should attain every value in $$\mathbb{C}$$ (except $$0$$) infinitely often in any neighborhood of $$z=0$$ ... there should be some points on the plot that show up, regardless of what PlotRange we choose.

So, my question is:

Why does ComplexPlot3D not work for $$f(z) = e^\frac{1}{z}$$? Is ComplexPlot3Dperhaps incompatible with essential singularities? Do I have some incorrect syntax or other mistake?

## closed as off-topic by Carl Woll, chuy, m_goldberg, Roman, MarcoBApr 19 at 3:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Carl Woll, chuy, m_goldberg, Roman, MarcoB
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I think the problem is your lowercase e. That symbol is undefined in Mathematica. Either use Esc ee Esc, E, or Exp. The following seems to work fine:
ComplexPlot[Exp[1/z], {z, -1 - I, 1 + I}]

• Thank you! I feel a bit silly knowing it's such a simple mistake ("$e$ is uppercase in mathematica"). That said, I do find it strange that $e$, $i$ (which are conventionally lowercase in pretty much all other math contexts) are uppercase in mathematica .. – Zubin Mukerjee Apr 18 at 18:06
• @ZubinMukerjee: Requiring E and I is totally consistent with the rule that all built-in Mathematica names begin with upper-case letters (or with \$ followed by an upper-case letter). The doubled-letter symbol you get with Esc ee Esc is just a shortcut for E. Of course you could avoid the whole issue by using the exponential function, as in Exp[1/z]. – murray Apr 18 at 20:36
• @murray But not exp[1/z] of course. :) – Kellen Myers Apr 18 at 21:49
• I use i as an iterator or an index all the time. IMO it’s good they stayed consistent with their naming conventions. – Chip Hurst Apr 19 at 0:38